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Trang chủ Giáo dục - Đào tạo Toán học Bồi dưỡng học sinh giỏi toán đại số giải tích 12 tập 1-lê hoành phò...

Tài liệu Bồi dưỡng học sinh giỏi toán đại số giải tích 12 tập 1-lê hoành phò

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ThS. L E H O A N H P H O Nhd gido Uu tu C c M i l B O I H O C D A I D U S S 8 N I N H O - G 7 G G I I O A I I T O T - Ddnh cho HS lop 12 on tap & nang cao kinang lam bai. - Chudn bi cho cdc ki thi quoc gia do Bo GD&DT to choc Mil NHA XUAT BAN DAI HQC QUOC GIA HA NQI A I N C H c Bin DUSNG , HQC SINH GO TOAN OAI SO -GIAI TICH Boi duQng hoc sinh gioi Toan Dai so 10-1. Boi duQng hoc sinh gioi Toan Dai so 10-2. - Boi duQng hoc sinh gioi Toan Hinh hoc 10. - Boi duOng hoc sinh gioi Toan Dai so 11. Boi duQng hoc sinh gioi Toan Hinh hoc 11. Bp de thi tif luan Toan hoc. Phan dang va pht/Ong phap giai Toan So phtfc. Phan dang va phucing phap giai Toan To hop va Xac suat. 1234 Bai tap tir luan dien hinh Dai so giai tich 1234 Bai tap ta iuan dien hinh Hinh hoc li/ong giac ThS. L E H O A N H Nha gido iCu tu B O I H O C D U S Q I N N H PHO G , G I O I T O A DAI SO-GIAI TICH 12 * - Danh cho HS lap 12 on rflp & ndng cao kfndng lam bai. - Chudn bj cho cdc ki thi qudc gia do Bo GD&DT td choc. Ha Npi NHA XUAT BAN DAI HQC QUOC GIA HA NQI N NHA XUAT BAN D A I HOC QUOC GIA HA N 0 I 16 Hang Chudi - Hai Ba Trirng Ha Npi Dien thoai: Bien tap-Che ban: (04) 39714896; Hanh chinh: (04) 39714899; Tdng bien tap: (04) 39714897 Fax: (04) 39714899 Chiu trach nhiem xuat bdn: Giam ddc PHUNG QUOC BAO Tong bien tap P H A M T H I T R A M Bien tap noi dung THUY NGAN Sda bdi NGOC H A N Che bdn CONG T I A N P H A Trinh bay bia SON K Y Ddi tdc lien ket xudt bdn CONG T I A N P H A SACH LIEN KET BOI DUONG HQC SINH GIOI TOAN DAI SO GIAI TICH 12 -TAP 1 Ma so: 1L-177DH2010 In 2.000 cuon, khd 16 x 24 cm tai cong ti TNHH In Bao bi Hung Phu So' xua't ban: 89-2010/CXB/11-03/DHQGHN, ngay 15/01/2010 j Quyet dinh xua't ban sd: 177LK-TN/XB In xong va nop ltiu chieu quy I I nam 2010. L d i N 6 I D A U De giup cho hoc sinh lap 12 co them tai lieu tu boi duong, ndng cao va ren luyen ki ndng gidi todn theo chuong trinh phdn ban mdi. Trung tdm sdch gido due ANPHA xin trdn trong giai thieu quy ban dong nghiep vd cdc em hoc sinh cuon: "Boi dudng hqc sinh gioi todn Dai so' Gidi tich 12 " nay. Cuon sdch nay nam trong bo sdch 6 cuon gom: - Boi ducmg hoc sinh gidi todn Hinh hoc 10. - Bdi ducmg hoc sinh gidi todn Dai so' 10. - Boi dudng hoc sinh gidi todn Hinh hoc 11. - Boi dudng hoc sinh gidi todn Dai so - Gidi tich 11. - Bdi dudng hoc sinh gidi todn Hinh hoc 12. - Boi dudng hoc sinh gidi todn Gidi tich 12. do nhd gido uu tu, Thac sTLe Hoanh Phd to'chirc bien soan. Ndi dung sdch duoc bien soan theo chuong trinh phdn ban: co bdn vd nang cao mdi ciia bd GD & DT, trong dd mot so van de duoc md rdng vdi cdc dang bdi tap hay vd kho dephuc vu cho cdc em yeu thich mud'n ndng cao todn hoc, cd dieu kien phdt trien tot nhat kha nang ciia minh. Cuon sdch la su ke thira nhung hieu bii't chuyen mdn vd kinh nghiem gidng day ciia chinh tdc gid trong qua trinh true tiep dirng ldp bdi dudng cho hoc sinh gidi cdc ldp chuyen todn. Vdi ndi dung sue tich, tdc gid da co'gang sap xep, chon loc cdc bai todn tieu bieu cho tirng the loai khdc nhau ung vdi ndi dung cua SGK. Mdt sd'bdi tap cd the khd nhung cdch gidi duqc dua tren nen tdng kien thuc vd ki nang co bdn. Hqc sinh can tu minh hoan thien cdc ki nang ciing nhu phdt trien tu duy qua viec gidi bdi tap cd trong sdch trudc khi ddi chieu vdi led gidi cd trong sdch nay, cd the mdt soldi gidi cd trong sdch con cd dong, hqc sinh cd thetu minh lam rd hon, chi tie't hon, ciing nhu tie minh dua ra nhung cdch lap ludn mdi hon. Chung tdi hy vong bd sdch nay se la mdt tdi lieu thie't thuc, bo ich cho ngudi day vd hqc, dqc biet cdc em hqc sinh yeu thich mdn todn vd hqc sinh chuan bi cho cdc ky thi qudc gia (tot nghiep THPT, tuyen sinh DH - CD) do bq GD & DT to chirc sap tdi. Trong qua trinh bien soqn, cudn sdch nay khdng the tranh khoi nhirng thieu sdt, chung tdi ra't mong nhdn duqc gop y ciia ban dqc gdn xa debq sdch hoan Men hon trong lah tdi ban. Moi y kien dong gop xin lien he: - Trung tam sach giao due Anpha 225C Nguyen Tri Phuong, P.9, Q.5, Tp. HCM. - Cong ti sach - thiet bj giao due Anpha 50 Nguyen Van Sang, Q. Tan Phii, Tp. HCM. DT: 08. 62676463, 38547464 . Email: [email protected] Xin chan thanh cam on! M U C L U C Chuong I : tTng dung dao ham de khao sat va ve do t h i cua ham so § 1. Tinh don dieu cua ham so 5 Dang 1: Dong bien, nghich bien, ham hang 5 Dang 2: Ung dung tinh don dieu 17 §2. Cue tri ciia ham so 37 Dang 1: Cue dai, cue tieu , 37 Dang 2: Ung dung ciia cue tri 48 §3. Gia tri Ion nhat va gia tri nho nhat 58 Dang 1: Tim gia tri ldn nhat, nho nhat 58 Dang' 2: Bai toan lap ham so 69 Dang 3: Ung dung vao phuong trinh 77 §4. Duong tiem can cua do thi ham so 88 Dang 1: Tim cac tiem can 88 Dang 2: Bai toan ve tiem can 96 §5. Khao sat va ve ham da thuc 103 Dang 1: Ham bac ha 104 Dang 2: Ham trung phuong 113 §6. Khao sat va ve ham hOu ti Dang 1: Ham so v = k (c * 0 va ad be * 0) cx + d 2 i Dang 2: Ham s6 v = (a * 0. a' * 0) a'x + b' §7. Bai toan thuong gap ve do thi a &X x + + 126 126 135 148 Dang 1: Tuong giao, khoang each, goc 149 Dang 2: Tiep tuyen. tiep xuc 159 Dang 3: Yeu to co dinh. doi xung - quy tich 170 Chirong I I : Ham so luy thua ham so mu va ham so logarit § 1. Quy tac bien doi va cac ham so 186 Dang 1: Bien doi luy thua - mu - logarit 188 Dang 2: Cac ham so mu. luy thua, logarit 200 Dang 3: Bat dang thuc va GTLN, GTNN 212 C H U O N G I:U N G DUNG D A O H A M OE SAT V A VE D O THj C U A H A M K H A O SO §1. TINH DON DIEU CUA HAM SO A. K I E N THLTC CO B A N • Dinh nghTa: Ham so f xac dinh tren K la mot khoang, doan hoac nira khoang. - f dong bien tren K neu vdi moi Xi, X2 6 K: X] < X2 => f(xi) < f(x2) - fnghich bien tren K neu vdi moi xi. xi e K: Xif(xi)>f(x2). • Dieu kien can de ham so don dieu Gia sir ham so co dao ham tren khoang (a; b) khi do: - Neu ham so f dong bien tren (a; b) thi f ' ( x ) > 0 vdi moi x e (a; b) - Neu ham so f nghich bien tren (a; b) thi f ' ( x ) < 0 vdi moi x e (a; b). • Dieu kien du de ham so don dieu - Gia sir ham so f co dao ham tren khoang (a; b) Neu f'(x) > 0 voi moi x e (a; b) thi ham so f dong bien tren (a; b) Neu f'(x) < 0 voi moi x e (a; b) thi ham so nghich bien tren (a; b) Neu f'(x) = 0 vdi moi x e (a; b) thi ham so f khong doi tren (a; b). - Gia sir ham so f co dao ham tren khoang (a; b) Neu f '(x) > 0 (hoac f '(x) < 0) vdi moi x e (a; b) va f '(x) = 0 chi tai mot so huu han diem cua (a; b) thi ham so dong bien (hoac nghich bien) tren khoang (a; b). B. P H A N D A N G T O A N DANG 1: B 6 N G B l i N , NGHICH BIEN, HAM HANG • Phuong phap xet tinh don dieu: - Tim tap xac dinh - Tinh dao ham, xet dau dao ham, lap bang bien thien - Ket luan Chii y: - Dau nhi thuc bac nhat: f(x) = ax + b, a ^ 0 -00 -b/a +co x f(x) trai dau a 0 ciing dau a - Dau tam thuc bac hai: f(x) = ax + bx + c, a * 0 Neu A < 0 thi f(x) luon ciing dau vdi a Neu.A = 0 thi f(x) luon cung dau vdi a, trir nghiem kep Neu A > 0 thi dau "trong trai - ngoai ciing" X -CO X] X 2 2 f(x) ciing dau a 0 trai dau a -BDHSG DSGT12/1- 0 ciing d i u a +00 V i du 1: Xet chieu bien thien ciia ham sd: a) y = x - 6x + 5 2x + x - 3 b)y= - x 3 c) y = x - 2x + x + 1 d) y = - x + 4x - 7x + 5 Giai a) Tap xac dinh D = R. Ta co y' = 2x - 6. 2 3 2 3 2 3 2 Cho y' = 0 » 2x - 6 = 0 » x = 3. Bang bien thien X —oo 3 y' - +00 0 + — y Vay ham so nghich bien tren (-oo; 3), dong bien tren (3; +oo). b) D = R. Ta cd y' = 4x - 4x + 1 = (2x - l ) > 0 v d i moi x 2 2 y' = 0 o x = —. Vay ham so dong bien tren R. 2 c) D = R . Ta co y' = 3x - 4x + 1 2 Cho y' = 0 o 3x - 4x + 1 = 0 <=> x = - hoac x = 1. 3 BBT X —00 +00 1/3 1 + + 0 0 y' 2 J ^ * y — ^ Vay ham so dong bien tren moi khoang (-co; —) va (1; +oo), nghich bien 3 tren khoang (—; 1). 3 d) D = R Ta cd y' = - 3 x + 8x - 7 V i A ' = 1 6 - 2 1 < 0 nen y' < 0 vdi moi x do do ham so nghich bien tren R. V i du 2: Xet chieu bidn thien cua cac ham so sau: a) y = x - 2x - 5 b) y = x + 8x + 9 Giai a) D = R. Ta co y' = 4x - 4x = 4x(x - 1) Cho y' = 0 <=> 4x(x - 1) = 0 <=> x = 0 hoac x = ±1 BBT +00 X —00 1 0 -1 2 4 2 4 3 2 2 2 y' - y ^ o •+ ^ * 0 ^ 0 + ^ Vay ham sd nghich bien tren moi khoang (-co; - 1 ) va (0; 1), ddng bidn tren moi khoang ( - 1 ; 0) va (1; +=»)• 6 -BDHSG DSGT12/1- b) D = R. Ta co y' = 4x + 16x = 4x(x + 4),y' = 0 o x = 0. y' > 0 tren khoang (0; +co) => y dong bien tren khoang (0; +co) y' < 0 tren khoang (-co; 0) => y nghich bien tren khoang (-co; 0). V i du 3: Xet su bien thien cua ham so: 1 3x-8 d)y c)y a) y = x + — b)y (x-4) 1-x x 3 2 Giai a) Tap xac dinh D = R \ {0} _3_ x - 3 , y' = 0 <» X Ta co y' = 1 ,.2 2 2 BBT: X -co +00 Q N / 3 + y' :V3" 0 - - 0 + y Vay ham so dong bien tren khoang (-co; - ^ 3 ) va ( J 3 ; +oo), nghich bien tren m6i khoang ( - S ; 0) va (0; V3 ). > 0 vdi moi x ^ 0 nen ham so dong bien b) D = R \ {0}. T a c d y ' = 1 tren moi khoang (-oo; 0) va (0; +co). -5 c) D = R \ { 1 } . Ta cd y' = - < 0 vdi moi x -t- 1 nen ham so nghich (1-x) bien trong cac khoang (-co; 1) va (1; +oo). 3 2 d) D = R\ {4}.Tacdy'= ———- (x-4) y' < 0 tren khoang (4; +co) nen y nghich bien tren khoang (4; +co). y' > 0 tren khoang (-co; 4) nen y ddng bien tren khoang (-co; 4). V i du 4: Tim cac khoang don dieu ciia ham so: , x-2 2x a) y = -5 b) y x -9 X +X +1 Giai - x + 4x + 3 a) D = R. Ta cd: y' (x + x + l ) 3 2 2 2 2 y' = 0 e> x - 4x - 3 = 0 <=> x = 2 ± ^7 BBT: X -co 2-V7 2+V7 2 - y' y -BDHSG DSGT12/1- — 0 ^ — + * 0 - +°o * 7 Vay ham so dong bien tren khoang (2 - yfl ; 2 + V7 ) va nghich bien tren cac khoang (-co; 2 - 77 ) va (2 + ; +oo). b) D = R\{-3;3}.Tac6y'=^^ < ,Vx*±3. (x - 9 ) Do do y' < 0 tren cac khoang (-co; - 3 ) , (-3; 3), (3; +oo) nen ham so da cho nghich bien tren cac khoang do. V i du 5: Xet su bien thien cua ham sd: 0 2 a) y = V 4 - x 2 b)y = Vx - 2 x + 3 2 2 c)y d)y Vl6-: x +2 Giai a) Dieu kien 4 - x > 0 < = > - 2 < x < 2 nen D = [-2; 2] 2 V d i - 2 < x < 2 thi y' BBT: X -2 n 0 + y' ^ y , y ' = 0<=>x = 0. V4~ 2 ^ — ^ Vay ham so dong bien tren khoang (-2; 0) va nghich bien tren khoang (0; 2). Do ham so f lien tuc tren doan [-2; 2] nen ham so dong bien tren doan [-2; 0] va nghich bien tren doan [0; 2]. b) V i A' = 1 - 3 < 0 nen x - 2x + 3 > 0, V x => D = R. •p , 1 2x-2 x - l , . , Ta co y = — == = . y = 0 o x = 1. 2vx -2x + 3 Vx -2x +3 y ' > 0 o x > l , y ' < 0 o x < l nen ham so nghich bien tren khoang (-co; 1) va dong bien tren khoang (1; +00). c) D K : 16 - x > 0 o x < 16 o - 4 < x < 4. D = (-4; 4). 2 2 2 2 Ta co v' = 2 1 > 0, Vx e (-4; 4). 6 (16-x )Vl6-x 2 2 Vay ham so dong bien tren khoang (-4; 4). d) D = [0; +00). V d i x > 0, y' = _, y' 2^y^(x + 2) 2 +00 X 2 BBT: X y y 0 + 0 ^ Vay ham sd ddng bien tren (0; 2) va nghich bien tren (2; +00). 8 -BDHSG DSGTu/1- V j du 6: Tim khoang don dieu cua ham so a) y = V x ( x - 3) b)y = - x c)y d)y = 7x -6 2 x +1 Vl-x Giai a) D = [0; +oo). V d i x > 0, ta cd: y 1 3NRX-1) ( x - 3 ) + vx = r 2Vx BBT: X ,y 2x 0 1 +GO 0 y' y 0<=>x= 1. + — — — ^ Vay ham so nghich bien tren khoang (0; 1) va dong bien tren khoang (i';+°o). b) D = R. V d i x ^ 0, ta co: y' = — 3 3vV y' = 0 <=> x = 1 <=> x = ± 1 . 3vV 2 y' > 0 o ^/x " > l < = > x > l c i > x < - l hoac x > 1. 2 2 y' < 0 %/x " < l e > x < l o - K x < l . Vay ham so dong bien tren cac khoang (-co; -1) va (1; +co), nghich bien tren khoang ( - 1 ; 1). c) Tap xac dinh D = (-co; -^6 ) U (x/6 ; +oo). 2 2 Tacd: y' = -^^ ^£L,y = 0»x = ±3. (x -6)v'x -6 1 7 2 BBT: X y' 2 —CO -3 + 3 V6 V6 0 - • y _ 0 +CO + • Vay ham so dong bien tren cac khoang (-co; - 3 ) va (3; +oo). nghich bien tren cac khoang (-3; - v o ) va ( v o ; 3). d) D = (-QO; 1). T a c d y ' = , ~ 2V(l-x) 3 > 0, V x < l . X 3 Vay ham sd dong bien tren khoang (-co: 1). -BDHSG DSGT12/1- V j du 7: Xet su bien thien cua ham sd: 3 a ) y - - - x + smx ) b y = x + c o g 2 x Giai 3 a) D = R. Ta cd y' = - - + cosx < 0, Vx nen ham sd nghich bien tren R. b) D = R. Ta cd y' = 1 - 2cosxsinx = 1 - sin2x y' = 0o sin2x = 1 <=> x = - +kit,keZ. 4 Ham sd lien tuc tren moi doan [- + kn; — + (k + 4 4 va y' > 0 tren moi khoang (- + kn; - + (k + 1)TI) nen ddng bien tren 4 4 moi doan [- + kn; - + (k + l)7tl, keZ. 4 4 ' Vay ham so dong bien tren R. Vi du 8: Tim khoang dong bien, nghich bien cua ham so: a) y = x - sinx tren [0; 2TT] b) y = x + 2cosx tren (0; n). Giai a) y' = 1 - cosx. Ta cd Vx [0; 2n] => y > 0 va y' = 0 <=> x = 0 hoac x = 2n. Vi ham so lien tuc tren doan [0; 2n] nen ham so ddng bien tren doan [0; 2n]. b) y' = 1 - 2 sinx. Tren khoang (0; 7t). v 1 1 y'>0o sinx <-<=> - < x < — 2 6 6 y' < 0 » sinx > - <=>0 V x = I x | > x, Vx nen f '(x) < 0, Vx do dd ham sd f nghich bien tren R. b) f'(x) = -2(sin2x+ 1)<0 vdi moi x. 2 2 10 -BDHSG DSGT12/1- f'(x) = 0 o s i n 2 x = -lc^>2x = - - + 2 k n o x = - - +kn,k& 2 4 Ham f(x) lien tuc tren moi doan [ - - + kn; ~ Z. + (k + \)n] va f'(x) < 0 tren moi khoang (-— +kn; — + (k+l)n) nen ham so nghich bien tren moi doan 4 4 [ - - + k ; r ; - - +(k + l)n], k e Z. 4 4 Vay ham sd nghich bien tren R. Cach khac: Ta chung minh ham sd f nghich bien tren R: VXj, x e R, x < x => f(Xj) > f(x ). That vay, lay hai sd a, b sao cho a < X| < X2 < b. Ta cd: f ' ( x ) = -2(sin2x + 1) < 0 vdi moi x e (a; b). V i f '(x) = 0 chi tai mot sd huu han diem cua khoang (a; b) nen ham sd f nghich bien tren khoang (a; b) => dpcm. V i du 10: Chung minh rang cac ham so sau day dong bien tren R. 2 x 2 2 a) f(x) = x - 6x + 17x + 4 b) f(x) = 2x - cosx + S sinx. Giai a) f ' ( x ) = 3x - 12x + 17. V i A' = 36 - 5 1 < 0 nen f ' ( x ) > 0 vdi moi x, do dd ham so dong bien tren R. 3 2 2 V3 b) y' = 2 + sinx - v3 cosx = 2(1 + — sinx cosx). 2 2 = 2[1 + sin(x - —)] > 0, vdi moi x. 3 Vay ham sd ddng bien tren R. V i d u l l : Chung minh ham so: x-2 a) y = ddng bien tren moi khoang xac dinh cua nd. x +2 - x — 2x + 3 b) y = nghich bien tren moi khoang xac dinh cua nd. x +1 Giai 4_ a) D = R \ { - 2 } . Ta cd y' = — > 0 vdi moi x * - 2 (x + 2) 2 2 Vay ham so dong bien tren moi khoang (-oo; - 2 ) va (-2; +oo). x -2x-5 b) D = R \ { - l } . T a c d y ' = ~ < 0 vdi moi x * - l (vi A' = 1 - 5 < 0). (x + 1) ' ' Vay ham so nghich bien tren mdi khoang (-oo; - 1 ) va ( - 1 ; +oo). V i du 12: Chung minh ham so: 2 2 -BDHSG DSGT12/1- v 1 1 a) y - i + dong bien trong khoang ( - 1 ; 1) va nghich bidn trong cac x 2 khoang (-co; -1) va (1; +oo). , . sin(x + a) , , _ , , ) y~ ( ^ b + krt; k e Z) don dieu Uong mdi khoang xac dinh. sin(x + b) • ° Giai , , l(l + x )-2x.x 1-x b a 2 2 = (i x ) ( T ^ ' Ta cd y' > 0 <=> 1 - x > 0 <=> - 1 < x < 1. y ' < 0 < = > l - x < 0 < = > x < - l hoac x > 1. Tir do suy ra dpcm. b) Ham sd gian doan tai cac diem x = -b + kn (k e Z). , _ sin(x + b) cos(x + a) - sin(x + a) cos(x + b) _ s i n ( b - a ) a ) y 2 2 = y = 0 o x = ± 1 + 2 2 sin (x + b) s m ( x + b) 2 2 (do a - b * kn) V i y' ?t 0 va y' lien tuc tai moi diem x * - b + kn, nen y' giu nguyen mot dau trong moi khoang xac dinh, do do ham so don dieu trong moi khoang do. V i du 13: Chung minh: a) sin x + cos x = 1, Vx. b) cosx + sinx. tan— = 1, Vx e (-— ; —). 2 Giai a) Xet f(x) = sin x + cos x, D - R. f '(x) = 2sinxcosx - 2cosxsinx = 0, Vx. Do dd f(x) la ham hang tren R nen f(x) = f(0) = 1. 2 2 2 4 4 2 b) Xet f(x) = cosx + sinx tan-, D = (-—; — ). 2 4 4 r-,/ x x sinx . x x 1 (x) = -smx + cosxtan— + = -sinx + cosx.tan— + tan— • 2cos 2 X X iX = -sinx + tan — (1 + cosx) = -sinx + tan— .cos — 2 2 2 —sinx + sinx = 0 voi moi x e (— ; —) 4 4 , , TT 71 Suy ra rang f la mdt ham hang tren khoang (-— ; — ). 2 2 2 2 Do dd f(x) = f(0) = 1 vdi moi x e (-—; - )• 4 4 V i du 14: Chung minh cac ham so sau la ham khong ddi -BDHSG DSGTU/1- a) f(x) = cos x + cos (x + —) - cosxcos(x + ^ ) 3 3 b) f(x) - 2- sin x - sm (a + x) - 2cosa.cosx.cos(a + x). Giai 2 2 2 2 a) f'(x) = -2cosxsinx - 2cos(x + - )sin(x + - ) + sinxcos(x + ^ ) + cosx.sin(x + ^ ) 3 3 o o o 71 7T = -sin2x - sin(2x + — ) + sin(2x + - ) = -sm2x - 2cos(2x + - ) . s i n 3 3 2 b = -sin2x - cos(2x + — ) = 0, vai moi x. 2 1 1 3 Do do f hang tren R nen f(x) = f(0) = 1 + = 6 w w 4 2 4 b) Dao ham theo bien x (a la hang so). f '(x) = -2sinxcosx - 2cos(a + x)sin(a + x) + 2cosa[sinxcos(a + x) + cosx.sin(a + x)]. = -2sin2x - sin(2x + 2a) + 2cosa.sin(2x + a) = 0. Do do f hang tren R nen f(x) = f(0) = 2 - sin a - 2cos a = sin a. V i du 15: Cho 2 da thuc P(x) va Q(x) thoa man: P'(x) = Q'(x) vdi moi x va P(0) = Q(0). Chung minh: P(x) = Q(x). Giai Xet ham so f(x) = P(x) - Q(x), D = R. Ta cd f '(x) = P'(x) - Q'(x) = 0 theo gia thiet, do do f(x) la ham hang nen f(x) = f(0) = P(0) - Q(0) = 0 vdi moi x. f(x) = 0 => P(x) ^ Q(x). V i du 16: Xac dinh ham so f(x) thoa man: f(0) = 8; f ( x ) . f '(x) = 1 - 2x (*). Giai 2 2 2 Ta cd (*) -(f (x))* = l-2xo (f (x))' = 3 - 6x. 3 Xet ham sd g(x) = f (x) - 3x + 3x thi g'(x) = ( f (x))' - 3 + 6x = 0. nen g(x) = C: hang so tren D, do do: f(x) - 3x + 3x = C ^> f (x) = - 3 x + 4x + C. 3 3 2 2 3 nen f(x) = N / - 3 X + 3x + C 2 V i f(0) = 8 => C = 64. 2 Vay f(x) = yj-3x + 3x + 64 , thu lai dung. V i du 17: T i m cac gia trj cua tham so a de ham so dong bien tren R. 2 a) f(x) = - x + ax + 4x + 3 • 3 3 2 b) f(x) = ax - 3x + 3x + 2 3 2 Giai a) f '(x) = x + 2ax + 4, A' = a - 4 - N6u a - 4 < 0 hay - 2 < a < 2 thi f '(x) > 0 vdi moi x e R nen ham so ddng bien tren R. 2 2 2 -BDHSG DSGT12/1- 13 - Neu a = 2 thi f '(x) = ( + 2) > 0 vai moi x * - 2 nen ham sd dong bien tren R. 2 x - Neu a = - 2 thi ham sd f '(x) = (x - 2) > 0 vdi moi x * 2 nen ham so dong bien tren R. 2 - Neu a < - 2 hoac a > 2 thi f '(x) = 0 cd hai nghiem phan biet nen f co doi dau: loai. 1 Vay ham sd ddng bien tren R khi va chi khi - 2 < a < 2. b) V (x) = 3ax - 6x + 3. 2 Xet a = 0 thi f '(x) = - 6 x + 3 cd doi dau: loai Xet a * 0, v i f khong phai la ham hang (y' = 0 tdi da 2 diem) tren dieu kien ham so dong bien tren R la f '(x) > 0, Vx (a>0 fa>0 f >0 <=> <^=> <=> « a > 1. [A'<0 [9-9a<0 |a>l Y l du 18: Tim cac gia tri cua m de ham sd nghich bidn tren R: a) f(x) mx - x b) f(x) = sinx - mx + C. Giai a) y' = m - 3x a 2 - Neu m < 0 thi y' < 0 vdi moi x e R nen f nghich bien tren R - Neu m = 0 thi y' = - 3 x < 0 vdi moi x e R, dang thuc chi xay ra vdi x = 0, nen ham so nghich bien tren R. 2 m ± ^ - Neu m > 0 thi y' = 0 o x = BBT X —00 Xl x 0 y' H y — +00 2 0 r » • Do do ham so dong bien tren khoang (xi, x ) : loai Vay ham so nghich bien tren R khi va chi khi m < 0. 2 b) V i f(x) khong la ham hang vdi moi m va C nen f(x) = sinx - m + C nghich bien tren R <=> f '(x) = cosx - m < 0, Vx a> m > cosx, Vx o m > 1. V i du 19: Tim m dd ham sd dong bien tren moi khoang xac dinh: (3m - l ) x - m + m by = x + 2 + m a) y x-l x+m Giai 2 a) D = R \ { - m } . Ta co: , _ (x + m)(3m - 1 ) - [(3m - l ) x - m + m 2 Y 14 = ' (x + m ) 2 4m -2m (x + m ) 2 2 -BDHSG DSGTU/l- Ham so dong bien tren moi khoang xac dinh <=> 4m - 2m > 0 2 <=> m < 0 hoac m > —. 2 m b) Ta cd y' = 1 vdi moi x * 1. (x-l) - Neu m < 0 thi y' > 0 vdi moi x * 1. Do do ham sd dong bien tren moi khoang (-oo; 1) va ( 1 ; +oo). XT' , x -2x + l - m - Neu m > 0 thi y = -„ (x-l) y' = 0 o x - 2 x + l - m = 0 < = > x = l + Vm BBT X —oo 1 —Vm 1 1 + Vm +oo 0 + + 0 y' 2 2 2 2 y Ham sd nghich bien tren moi khoang (1 - Vm ; 1) va ( 1 ; 1 + Vm ): loai. Vay ham sd ddng bien tren moi khoang xac. dinh cua nd khi va chi khi m<0. V i du 20: Tim a de ham sd: a) f(x) = x - ax + x + 7 nghich bien tren khoang ( 1 ; 2) 3 2 b) f(x) = — x - — (1 + 2cosa)x + 2xcosa + 1, a e (0; 2rr) dong bien tren 3 2 khoang (1; +oo). Giai a) f'(x) = 3 x - 2 a x + 1 Ham so nghich bien tren khoang ( 1 ; 2) khi va chi khi y < 0 vdi moi x e (1;2) ff(l) < 0 f 4-2a<0 13 <=> < <=>< <=>a>— [f(2)<0 [l3-4a<0 4 b) y' = x - (1 + 2cosa)x + 2cosa. Ta cd 0 < a < 2TT. y' = 0 o x = 1 hoac x = 2cosa. V i y' > 0 d ngoai khoang nghiem nen ham so dong bien vdi moi x > 1 khi 3 2 2 2 va chi khi 2cosa < 1 cosa < — o — < a < — 2 3 3 V i du 21: Tim m de ham sd y = (m - 3)x - (2m + l)cosx nghich bien tren R. Giai: y' = m - 3 + (2m - l)smx Ham so y khdng la ham hang nen y nghich bien tren R: y' ^ 0, Vx « m - 3 + (2m - l)smx < 0, Vx D5t t = sinx, - 1 < t < 1 thi m - 3 + (2m - l)smx = m - 3 + (2m - l ) t = f(t) -BDHSG DSGT12/1- 15 Dieu kien tuang duong: f(t) < 0, Vt e [-;1 1] [f(-l)<0 f-m-4<0 '9 °lf(l)^0 °l3m-2^0 ^ - ^ 4 ^ m V» du 22: Tim m de ham sd y = x + 3x + mx + m chi nghich bidn tren mpt doan cd dp dai bang 3. Giai: D = R, y' = 3x + 6x + m, A' = 9 - 3m Xet A' < 0 thi y' > 0, Vx: Ham luon dong bien (loai) 3 2 2 Xet A' > 0 <=> m < 0 thi y' = 0 cd 2 nghiem x,, x nen x, + x = -2, X]X = — 2 BBT: —CO x *1 0 - + y' y x 0 2 2 +00 2 3 + ^ ^ Theo de bai: x - X] = 3 o ( x - x , ) = 9 o 2 2 2 x +x -2x x = 9 2 2 t 2 4 15 <=> (x + x ) - 4 x x = 9 c i > 4 — m = 9 o m = (thoa) 3 4 V i du 23: Tuy theo tham &6 m, xet su bien thien cua ham sd: \ 1 3 ? , rx ix 2x + m a) y = - x - 2mx + 9x - m b) y = 3 x - l Giai a) D = R. Ta cd y' = x - 4mx + 9; A' = 4 m - 9 - Neu A' < 0 I m [ < — thi y' > 0, Vx nen ham so dong 2 2 t t 3 2 2 2 2 2 bien tren R. - Neu A' > 0 co 4 m > 9 co I m | > - thi y' = 0 cd 2 nghiem phan biet 2 xi, = 2m +V4m -9 Lap bang bien thien thi ham ddng bien tren 2 2 khoang (2m - V ^ m - 9 ; 2m + V 4 m - 9 ) va nghich bien tren m6i 2 2 khoang (-00; 2m - \/4m - 9 ) va (2m + V4m - 9 ; +00). 2 2 b)D = R\ {l}.Tacd y'= ~ ~"! 2 (x-l) - Neu m = - 2 thi y = 1, Vx * 1 la ham sd khong doi. - Neu m > - 2 thi y' < 0, Vx *1 nen ham so nghich bien tren moi khoang (-00; 1) va (1; +00). 2 - Neu m < - 2 thi y' > 0, Vx * 1 nen ham sd ddng bien tren moi khoang (-co; 1) va (1; +00). 16 -BDHSG DSGT12/1- DANG 2: UNG DUNG T I N H BON DI$U - Giai phirong t r i n h , he phirong trinh, bat phuong t r i n h : Xet f(x) la ham so v6 trai, neu can thi bien doi, chpn xet ham, dat an phu, .... Tinh dao ham rdi xet tinh don dieu. Neu ham sd f don dieu tren K thi phuong trinh f(x) = 0 cd toi da 1 nghiem. Neu f(a) = 0, a thupc K thi x = a la nghiem duy nhat cua phuong trinh f(x) = 0. Neu f cd dao ham cap 2 khdng ddi dau thi f la ham don dieu nen phuong trinh f(x) = 0 cd tdi da 2 nghiem. Neu f(a) = 0 va f(b) = 0 vdi a * b thi phuong trinh f(x)=0 chi cd 2 nghiem la x = a va x = b . - Chiing minh bat dang thiic: Neu y = f(x) cd y' > 0 thi f(x) dong bien: x > a => f(x) > f(a); x < b =>f(x) 0 thi y' dong bien tir do ta cd danh gia f '(x) rdi f(x),... V i du 1: Giai phuong trinh: vo - x + x - 72 + x - x = 1 Giai Dat t = x - x thi phuong trinh trd thanh: 7 3 + t - 7 2 - t = 1 , - 3 < t < 2. 2 2 2 Xet ham sd f(t) = 73 + t - 7 2 - t , -3 < t < 2. Vdi -3 < t < 2 thi f'(t) = + . > 0 nen f dong bien tren (-3; 2). 273 + t 272 - t Ta cd f ( l ) = 2 - 1 = 1 nen phuong trinh: 1 f(t) = f( 1) <=> t 1 o x — x — 1 = 0 <=> x l^H. = 2 = V i du 2: Giai phuong trinh 72x + 3 x +6x + 16 = 273 + 7 4 - x Giai: Dieu kien xac djnh: 3 2 f 2 x + 3 x + 6 x + 16>0 f(x + 2 ) ( 2 x - x + 8 ) > 0 <=>C cs> - 2 < x < 4 4-x>0 4-x>0 Phuong trinh tuong duong 72x + 3x + 5x +16 - 74 - x = 273 3 2 2 3 Xet ham s6 f(x) = N/2X 2 + 3 x + 6x + 16 - 74 - x .-2 < x < 4 2 3 ™, . 3(x + x + l ) . x Thi f (x) = —. = +— > 0 nen i dong bien ma 7 2 x + 3 x + 6 x + 16 274-x f ( l ) = 273 , do do phuong trinh trd thanh f(x) = f ( l ) o x = 1 Vdy phuong trinh co nghiem duy nhat x = l 2 c u 3 -BDHSG DSGT12/1- 2 17 V i du 3: Giai phucmg trinh yjx - 1 = Vx - 2 - x. Giai Dieu kien: x >%/2 2 Ta cd: V x - 2 = x + V x - 1 3 2 3 >x>l=>x >3=i>x>v 3 3 / Chia 2 ve cho vo? thi phuong trinh: ~ a. 1 x .vx x vx 2 Vx 4 V xVx Xet f(x) la ham sd ve trai, x > tfi thi ,, 9 5x X f'(x)= — 2x .Vx 2xVx x r , 3 <0. 2 2x Vx Do do ham so f nghich bien tren khoang ( y 3 ; +oo) ma f(3) = 0 nen phuong trinh cd nghiem duy nhat x = 3. B 5 r n 2 1 2x-5 Vi du 4: Giai phuong trinh: 3x - 18x + 24 2 1 x - l Giai 5 Dieu kien x * 1; —, phuong trinh trd thanh: 2 ( 2 x - 5 ) - - J _ = (x-l) |2x-5| |x-l| 2 2 Xet f(t) = t - i vdi t > 0. 2 Ta co: f '(t) = 2t > 0 nen f dong bien tren (0; +oo) Phuongtrinh:f(|2x-5|) = f(|x - l|)o 12x- 51 = |x-l| <=> 4x - 20x + 25 = x - 2x + 1 <=> 3x - 18x + 24 = 0. c ^ x - 6 x + 8 = 0 c o x = 2 hoac x = 4 (chon) V i du 5: Giai bat phuong trinh: 4 | 2x - 11 (x - x + 1) > x - 6x + 15x - 14 Giai BPT: | 2x - 11 .[(2x - l ) + 3] > (x - 2) + 3x - 6 (x - 2) + 3(x - 2) Xet ham sd f(t) = t + 3t, D = R. Ta cd f '(t) = 3t + 2 > 0 nen f dong bien tren R. BPT: f( | 2x - 11) > f(x - 2) o I 2x - 11 > x - 2. Xet x - 2 < 0 thi BPT nghiem dung. Xet x - 2 > 0 thi 2x - 1 > 0 nen BPT o 2 x - 1 > X - 2 < = > Vay tap nghiem la S = R. 2 2 2 2 2 3 2 3 2 3 3 3 2 18 X > - 1 : £) ng U -BDHSG DSGT12/1-
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